Nonlinear finite elements/Stress and strain in one and two dimension

Stresses and strain in one dimension

1-D Strain Measures

• Engineering strain:

$\varepsilon_{E} := \cfrac{l-L}{L} = \cfrac{\Delta L}{L}$

• Natural/Logarithmic/True strain:

$\varepsilon_{L} := \int_L^l \cfrac{dl}{l} = \ln(l)\Big|_L^l = \ln\left(\cfrac{l}{L}\right)$

Relation between engineering and true strain:

$\varepsilon_{L} = \ln\left(\cfrac{L+\Delta L}{L}\right) = \ln\left(1+\cfrac{\Delta L}{L}\right) = \ln\left(1+\varepsilon_{E}\right)$

• Green (Lagrangian) strain:

$\varepsilon_{G} := \frac{1}{2}\left(\cfrac{l^2-L^2}{L^2}\right) =\frac{1}{2}\left(\cfrac{l^2}{L^2}-1\right)$

• Almansi-Hamel (Eulerian) strain:

$\varepsilon_{A} := \frac{1}{2}\left(\cfrac{l^2-L^2}{l^2}\right) =\frac{1}{2}\left(1-\cfrac{L^2}{l^2}\right)$

1-D Stress Measures

• Engineering/Nominal stress:

$P = \sigma_{E} := \cfrac{T}{A}$

• Cauchy/True stress:

$\sigma = \sigma_{T} := \cfrac{T}{a}$

Relation between engineering and true stress (no volume change):

$\sigma_{T} = \cfrac{T}{a} = \cfrac{Tl}{AL} = \sigma_{E}\left(\cfrac{L+\Delta L}{L}\right) = \sigma_{E}\left(1 +\varepsilon_{E}\right)$

1-D Stress-Strain Relations

• True stress - Green strain:

$\sigma_{T} = E_{TG} \left(\cfrac{l^2 - L^2}{2 L^2}\right)$

• True stress - True strain:

$\sigma_{T} = E_{TT} \ln\left(\cfrac{l}{L}\right)$

Example

Stretching of a bar

$l^2 = D^2 + x^2 \implies \cfrac{dl}{dx} = \cfrac{x}{l}$

$\sin\theta = \cfrac{x}{l(x)}$

• Assume incompressible material.

$V = v ~;~~ AL = al$or :$V = al~;~~ a = V/l$}

• Equilibrium.

$T(x) = F$:$\implies~~ R(x) = T(x) - F = 0$

where

$\begin{align}T(x) & = \sigma(x)~a(x)~\sin\theta \\ & = \cfrac{\sigma(x)~a(x)~x}{l(x)} \\ & = {\cfrac{\sigma(x)~V~x}{l(x)^2}}\end{align}$

Stress-Strain relation 1

$\sigma(x) = E \left(\cfrac{l(x)^2 - L^2}{2 L^2}\right)$

Then,

$\begin{align} T(x) & = \cfrac{E~a(x)~x}{l(x)} \left(\cfrac{l(x)^2 - L^2}{2 L^2}\right)\\ & = \cfrac{E~V~x}{l^2} \left(\cfrac{l^2 - L^2}{2 L^2}\right) \end{align}$

and

$R(x) =\cfrac{E~V~x}{l^2} \left(\cfrac{l^2 - L^2}{2 L^2}\right) - F$

Highly nonlinear in $x$.

Stiffness

Stiffness = change in equilibrium equation due to change in position.

$K(x) = \cfrac{dR(x)}{dx} = \cfrac{dT(x)}{dx} ~~(\text{if}~ F ~\text{ is constant})$

Now,

$T(x) = \cfrac{\sigma(x)~V~x}{l(x)^2}$

Therefore,

$\begin{align} \cfrac{dT}{dx} & = V\left[x~\cfrac{d}{dx}\left(\cfrac{\sigma}{l^2}\right) + \cfrac{\sigma}{l^2}\right] \\ & = V\left[x~\cfrac{d}{d\sigma}\left(\cfrac{\sigma}{l^2}\right) \cfrac{d\sigma}{dx} + x~\cfrac{d}{dl}\left(\cfrac{\sigma}{l^2}\right) \cfrac{dl}{dx} + \cfrac{\sigma}{l^2}\right] = V\left[\cfrac{x}{l^2}\cfrac{d\sigma}{dl}\cfrac{dl}{dx} - \cfrac{2x\sigma}{l^3}\cfrac{dl}{dx} + \cfrac{\sigma}{l^2}\right] \\ & = \cfrac{Vx}{l^2}\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{dl}{dx} + \cfrac{V\sigma}{l^2} = \cfrac{Vx}{l^2}\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x}{l} + \cfrac{V\sigma}{l^2} \\ \implies K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \end{align}$

$\sigma= E \left(\cfrac{l^2 - L^2}{2 L^2}\right) \implies \cfrac{d\sigma}{dl}= \cfrac{E~l}{L^2}$

$\begin{align} K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = a\left(\cfrac{E~l}{L^2} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = \cfrac{A}{L}\left(E - 2S\right)\cfrac{x^2}{l^2} + \cfrac{S~A}{L} \end{align}$

Initial stress/Geometric stiffness

$\cfrac{S~A}{L} ~;~~ S= \sigma \cfrac{L^2}{l^2} = P \cfrac{L}{l}$

Stress-Strain relation 2

$\sigma(x) = E \ln\left(\cfrac{l(x)}{L}\right)$

Then,

$\begin{align} T(x) & = \cfrac{E~a(x)~x}{l(x)} \ln\left(\cfrac{l(x)}{L}\right) \\ & = \cfrac{E~V~x}{l^2} \ln\left(\cfrac{l}{L}\right) \end{align}$

and

$R(x) =\cfrac{E~V~x}{l^2} \ln\left(\cfrac{l}{L}\right) - F$

Highly nonlinear in $x$.

$\sigma= E \ln\left(\cfrac{l}{L}\right) \implies \cfrac{d\sigma}{dl}= \cfrac{E}{l}$

Stiffness

$\begin{align} K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = a\left(\cfrac{E}{l} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = \cfrac{a}{l}\left(E - 2\sigma\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \end{align}$

Initial stress/Geometric stiffness:

$\cfrac{\sigma~a}{l}$

Strain Measures in two dimensions

Strains in two dimensions

Small strains

$\begin{align} \varepsilon_{xx} & = \frac{\partial u_x}{\partial x} \\ \varepsilon_{yy} & = \frac{\partial u_y}{\partial y} \\ \varepsilon_{xy} & = \frac{1}{2}\left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right) \end{align}$

For 90$^o$ rotation,

$u_x = -Y - X ~;~~ u_y = X - Y$

Then strains are:

$\begin{align} \varepsilon_{xx} & = -1 \\ \varepsilon_{yy} & = -1 \\ \varepsilon_{xy} & = 0 \end{align}$

Rotation should not lead to non-zero strains!

Finite strains

Green strain in two dimensions

For 90$^o$ rotation,

$u_x = -Y - X ~;~~ u_y = X - Y$

Then,

$E_{xx} = 0~;~~ E_{yy} = 0~;~~E_{xy} = 0$

Green strain (1-D)

$\varepsilon_{G} = \cfrac{l^2-L^2}{2L^2}$

In 2-D:

$E_{xx} = \cfrac{ds^2-dX^2}{2dX^2}$

Now,

$ds^2= dX^2\left(1 +\frac{\partial u_x}{\partial X}\right)^2 + dX^2\left(\frac{\partial u_y}{\partial X}\right)^2$

Therefore,

$\begin{align} E_{xx} & = \cfrac{dX^2}{2dX^2}\left[ \left(1 +\frac{\partial u_x}{\partial X}\right)^2 + \left(\frac{\partial u_y}{\partial X}\right)^2 - 1 \right] \\ & = \frac{\partial u_x}{\partial X} + \frac{1}{2}\left[ \left(\frac{\partial u_x}{\partial X}\right)^2 + \left(\frac{\partial u_y}{\partial X}\right)^2 \right] \end{align}$

Similar for $E_{yy}$ and $E_{xy}$.